Question about velocity/position vectors?

Note: since velocity is a function of time, you need to do integration with respect to time to find the position as a function of time

V(t) = Vx i + Vy j

Vx = x component of velocity = 2.4 - 1.6t^2

Vy = y component of velocity = 4t

Vx = dX/dt

Vy = dY/dt

X(t) = Integral[Vxdt] = Integral[(2.4 - 1.6t^2)dt] = 2.4t - 1.6/3 t^3 + X(0)

Y(t) = Integral[Vydt] = Integral[(4t)dt] = 2t^2 + Y(0)

At t = 0 the bird is at the origin (0,0)

so X(t=0) = X(0) = 0

and Y(t=0) = Y(0) = 0

so

X(t) = 2.4t - 1.6/3 t^3; Y(t) = 2t^2 are the equation of the position vector of the bird in the XY plane as a function of time

What is the bird's altitude (y-coordinate) as it flies over x = 0 for the first time after t = 0?

find out at what value of t, X(t) = 0 after t = 0 meaning t > 0

X(t) = 0

2.4t -1.6/3 t^3 = 0

(2.4 - 1.6/3 t^2)t = 0 which implies t=0 or t^2 = 2.4 * 3/1.6

we are interested in t > 0

so t^2 = 2.4*3/1.6

t = 2.12

Plug t=2.12 in Y(t)

Y(t=2.12) = 9

answer, bird's altitude = 9

the velocity is the derivative of position

so we will integrate the velocity to see where the bird is at any given time

p = i(2.4t - 0.53*t^3 + c1) + j(2t^2 + c2)

solve for c1 and c2 with t=0 x=y=0, they turn out to be 0

p = i(2.4t - 0.53*t^3) + j(2t^2)

so we find when it crosses x=0

2.4t - 0.53t^3 = 0

0.53t^3 = 2.4t

t^2 = 4.528

t = 2.123 (or -2.123 but we cant have a negative time)

so

y = 2t^2

y = 2(2.123)^2

y = 9.056

I didn't check my work but I'm pretty sure that's how you go about it

the area vector r is given by using the critical of the cost. r = (2.4t - a million.6t^3/3)i + 2t^2 j It x coordinate is 0 while (2.4t - a million.6t^3/3) = 0 you may take it from right here to sparkling up for t and plug that throughout the time of for the j part of r to discover the altitude.